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Old 09-03-2002, 09:53 AM   #1 (permalink)
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What is a 9 Percent Road Grade in Degrees?

Drove from Harrisonburg, VA to Seneca Rocks, WV on Route 33 west, the steepest, windiest (sp?) primary road I have ever driven on. It has many 7 and 9 percent grades and 20 mph curves.

How does the percent translate to degrees???????????

Thanks,

:D

BTW, I spotted a MY Z06 at 2:45PM on Saturday (Aug 31) afternoon leaving the Seneca Rocks Visitors Center on route 33. Any idea who that was?? Didn't get close enough to see a license plate.
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Old 09-03-2002, 10:18 AM   #2 (permalink)
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9/100=x/90 x= 9x90/100=8.1

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Old 09-03-2002, 10:21 AM   #3 (permalink)
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Great road! Used it for my shakedown cruise in June. I'd prefer somebody with civil engineering background directly answer your question.
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Old 09-03-2002, 11:38 AM   #4 (permalink)
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I believe 9% grade means 9 feet up for every 100 feet horizontal. So the grade in degrees is the inverse tangent of .09. Mr. Hewlett and Packard say this is 5.14 degrees.
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Old 09-03-2002, 11:46 AM   #5 (permalink)
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.09% or 0 degrees 5 minutes 24 seconds
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Old 09-03-2002, 11:53 AM   #6 (permalink)
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Wrong key
9%=.09 degrees or 0 degrees 5 minutes 24 seconds
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Old 09-03-2002, 12:01 PM   #7 (permalink)
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Rosso: I believe when talking about roads, they mean 9% of vertical (or 90%) or, like FlyingZ06 said.... .09 X 90= 8.1 degrees, or if downhill, -8.1 degrees.

In other words, a 25% grade would be
22.5 degrees or the roof of an average house(6/12 pitch) and a 50% grade would be 45 degrees or the roof of a Cape Cod House (12/12 pitch)


Joe

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Old 09-03-2002, 12:51 PM   #8 (permalink)
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The correct answer is: about 5 degrees. As stated above, 9% grade means 9 feet of elevation per 100 feet of horizontal distance. Doesn't seem like much, but to a wheeled vehicle it makes a huge difference.

Pitch on a roof is measured similarly, like embankment slopes, but is not directly the same as percent grade. Most embankments are a slope of 2:1, or less (for every two feet you go horizontally, elevation changes by 1 foot - or about 26.5 degrees). Now, grade is actually the change in elevation change divided by a measured horizontal distance, expressed as a percent. In the case above, 9 feet divided by 100 feet (.09 or 9%). Now, a 12:12 pitch on a roof means 12 inches of elevation for every 12 inches of horizontal distance. This does give a 45 degree angle, from horizontal, for the roof. To duplicate the same slope on a road (for percent grade) means that for every 100 feet you travel horizontally, you also change elevation by 100 feet . In percentages, this would be a 100 percent grade (100/100).

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Old 09-03-2002, 02:13 PM   #9 (permalink)
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Yeah!
^
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What he said...lol

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Old 09-03-2002, 02:21 PM   #10 (permalink)
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Thanks for all the input.

As stated by FlyNavy and several engineers I work with, the grade in degrees is the inverse tangent (arc tangent) of .09 or 5.14 degrees. For you HP users that's .09 [enter], [blue key], then Tangent key (Tan to the negative 1).

It sure feels like a lot more than 5 degrees!

However, I used to sprint race at the velodrome near Bellevue, WA. It had 30 degree banks in the turns and you could hardly stand up on it. I believe Daytona also has 30 degree banks.



Case closed...

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Old 09-03-2002, 02:45 PM   #11 (permalink)
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From an engineering standpoint I understand how your calculations are derived. 10 feet down in a hundred foot run is expressed as 10% grade.(run vs. rise)

When I measure a scene for accident reconstruction, or a race track measures the grade (actually the super elevation of the track) what is measured is the actual angle (from horizontal) of the surface.
If I use an inclinometer on a 12/12 roof, I get 45 degrees, but from an engineering standpoint that would be considered a 100% grade. Is this right?

Percent of grade in roadways signs is the percent of the ratio??? , not the percent from level. Do I have it right now?

Joe

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Old 09-03-2002, 03:15 PM   #12 (permalink)
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My head is hurts.

There has to be a simple formula that can be applied.
Won't somebody give it up.
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Old 09-03-2002, 03:27 PM   #13 (permalink)
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Rosso: If I understand this right now (thanks to John and Donna above), your road sign, if applied to Daytona (driving from the top to the bottom) would have to state (approx) 57% downgrade.

Okay, no more math for a while.

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Old 09-03-2002, 03:45 PM   #14 (permalink)
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Quote:
Originally posted by FiveOJoe

If I use an inclinometer on a 12/12 roof, I get 45 degrees, but from an engineering standpoint that would be considered a 100% grade. Is this right?

So: If Daytona is 100 foot wide in the turns, the inside is 30 feet lower than the outside, or is the track actually banked at 30 degrees.

Percent of grade in roadways signs is the percent of the ratio??? , not the percent from level. Do I have it right now?

Joe
Yes, 45 degrees equals a 100% grade.

If your dimensions are correct for Daytona, the "grade" of the curves is 30%. This translates to an angle of 16.7 degrees. Now, if the "bank" is 30 degrees and the track width is 100 feet (this is important- is it the width of the inclined surface or the horizontal width- I am assuming the inclined surface is 100 feet), the elevation difference would be 50 feet.

You can think of grade as the percent of the ratio, which is also the percent from level. You are comparing how "fast" the elevation changes relative to the horizontal change. In this way, a slope greater than 45 degrees gives you a percentage greater than 100%; this means that the elevation is changing "faster" than the horizontal - 130% grade equals 52 degrees. At 45 degrees the elevation and horizontal are changing at the same "speed"

SAL:

the "formula" requires the use of a Scientific Calculator. Simply enter in the slope, in decimal form (9% would be .09) and then press the "inverse tangent" key; this is symbolized as ATAN, ARCTAN, or TAN with exponent "-1". You should have to hit a "shift" key to access this function - the regular button will say TAN (or tangent); "shift" then this button is "inverse tangent". The result displayed will be the angle measure.
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Old 09-03-2002, 06:43 PM   #15 (permalink)
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Don't know the math, but 9% is steep. I would hate to have a big truck on that hill.
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